$$\left|\frac{x^2} {x-1}\right|\leq 1$$

In the case where $x>1$, there are no real roots.

So in the case where $x<1$:

I opened the L.H.S. with a $-$ sign (as the modulus function has to remain positive). After this I multiplied both sides of the inequation by $-1$ giving me: $\frac{x^2}{x-1} \geq -1$ [The inequality sign flips due to multiplication with a negative number].

What have I done wrong, as doing this led me to the wrong answer...

  • 1
    $\begingroup$ You did nothing wrong so far. $\endgroup$ – mathlove Dec 29 '16 at 10:08

As you said $x>1$ is not possible so suppose $x<1$ then

$$ \left|\frac{x^2}{1-x}\right|=\frac{x^2}{1-x}\leq 1 $$

So if you rearrange this you get $x^2+x-1\leq 0$. Using quadratic formula you get $\frac{-1-1\sqrt{5}}{2}<x<\frac{-1+\sqrt{5}}{2}$

With what you are doing steps are following

$$ \frac{x^2}{x-1}\geq -1 $$

Multiply both sides with $x-1<0$ then inequality change its sign again and

$$ x^2\leq -x+1 $$ which yields the same.

  • $\begingroup$ Thanks a lot! I realized the mistake. $\endgroup$ – Harsha G. Jan 1 '17 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.